The molar heats of fusion and sublimation of molecular
iodineare 15.27 kj/mol and 62.30 kj/mol, respectively. estimate the
molarheat of vaporization of liquid iodine.

47.03 KJ/mol

Explanation:

given:

molar heat of fusion, Hfusion = 15.27 KJ/mol

molar heat of sublimation, Hsublime = 62.30 KJ/mol

molar heat of vaporisation, Hvaporise = ?

Hsublime = Hfusion + Hvaporise

Hvaporise = Hsublime - Hfusion

Hence, the molar heat of vaporization of liquid iodine

= 62.30 - 15.27 = 47.03 KJ/mol

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons